Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007ll
#define INF 1000000000ll
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v) sort(v.begin(),v.end());v.erase(unique(v.begin(),v.end()),v.end());
#define pb push_back

vector<ll> dijkstra(ll s, vector<vector<P>>& adj) {
	vector<ll> dir(adj.size(),INF*INF);
	dir[s]=0;
	priority_queue<P,vector<P>,greater<P>> q;
	q.push(P(0,s));
	while(!q.empty()) {
		ll d=q.top().first;
		ll p=q.top().second;
		q.pop();
		if(dir[p]!=d) continue;
		REP(i,adj[p].size()) {
			ll c=adj[p][i].second;
			ll np=adj[p][i].first;
			if(dir[np]>d+c) {
				dir[np]=d+c;
				q.push(P(dir[np],np));
			}
		}
	}
	return dir;
}


int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n,m,s,t;
	cin>>n>>m>>s>>t;
	s--; t--;
	vector<vector<P>> adj(n);
	REP(i,m) {
		ll x,y;
		cin>>x>>y;
		x--;
		y--;
		adj[x].pb(P(y,1));
		adj[y].pb(P(x,1));
	}
	vector<ll> froms=dijkstra(s,adj);
	vector<ll> fromt=dijkstra(t,adj);
	map<ll,ll> cnt;
	REP(i,n) cnt[fromt[i]]++;
	ll ans=0;
	REP(i,n) ans+=cnt[froms[t]-froms[i]-2];
	cout<<ans<<endl;
}